Which of the following numbers is a multiple of 9? ${48,51,78,99,100}$
The multiples of $9$ are $9$ $18$ $27$ $36$ ..... In general, any number that leaves no remainder when divided by $9$ is considered a multiple of $9$ We can start by dividing each of our answer choices by $9$ $48 \div 9 = 5\text{ R }3$ $51 \div 9 = 5\text{ R }6$ $78 \div 9 = 8\text{ R }6$ $99 \div 9 = 11$ $100 \div 9 = 11\text{ R }1$ The only answer choice that leaves no remainder after the division is $99$ $ 11$ $9$ $99$ We can check our answer by looking at the prime factorization of both numbers. Notice that the prime factors of $9$ are contained within the prime factors of $99$ $99 = 3\times3\times11 9 = 3\times3$ Therefore the only multiple of $9$ out of our choices is $99$. We can say that $99$ is divisible by $9$.